# Shape of the state space under different tensor products

I am currently studying generalized probabilistic theories. Let me roughly recall how such a theory looks like (you can skip this and go to "My question" if you are familiar with this).

Recall: In a generalized probabilistic theory (GPT), the set of states is given by a convex subset $\Omega \subset V$ of a real vector space (in the special case of quantum theory, $\Omega$ is the set of density operators and $V$ is the vector space of Hermitian operators on a Hilbert space). A the statistics of a measurement are described by a set of effects. An effect is a linear functional $e \in V^*$ such that $0 \leq e(\omega) \leq 1$ for all $\omega \in \Omega$. Let the set of effects be denoted by $E(\Omega)$. The unit effect $u \in E(\Omega)$ is given by $u(\omega) = 1$ for all $\omega \in \Omega$ (in a GPT, the set of states $\Omega$ is always such that such a functional $u \in V^*$ exists). A measurement is a set $\{ e_1, \ldots, e_n \}$ of effects such that $\sum_{i=1}^n e_i = u$.

In order to describe composite systems in a GPT, the concept of tensor products is introduced (in what follows, a "tensor product" of state spaces $\Omega_A$ and $\Omega_B$ is a rule that tells you how to combine systems, and this rule does not have to coincide with what is usually called a tensor product in mathematics, whereas the tensor product $e_A \otimes e_B$ means the usual tensor product; I think this is a bad terminology, but it is very common in the theory of GPTs). A state $\omega^{AB} \in \Omega^{AB}$ of a system composed of two subsystems $A$ and $B$ has to satisfy $$\text{normalization:} \quad (u^A \otimes u^B)(\omega^{AB}) = 1$$ and $$\text{positivity:} \quad (e_A \otimes e_B)(\omega^{AB}) \geq 0 \quad \forall e_A \in E(\Omega_A), \forall e_B \in E(\Omega_B),$$ where $\otimes$ denotes the usual (in mathematical terminology) tensor product. These two requirements have to be fulfilled by every state of a composite system; they are the minimal restrictions for a composite system.

The maximal tensor product $\Omega_A \otimes_\text{max} \Omega_B$ of two systems is given by the set of all $\omega^{AB} \in V_A \otimes V_B$ that satisfy normalization and positivity (it's called maximal since minimal restrictions lead to a maximally large set of states).

The other extreme case is the minimal tensor product which is given by all convex combinations of product states $\omega_A \otimes \omega_B$, i.e. by all mixtures of product states.

There are also other possible ways to combine systems, i.e. other tensor products apart from the maximal and the minimal tensor product. For example, the "tensor product" in the quantum case (encompassing all density operators on $\mathcal{H}_A \otimes \mathcal{H}_B$) is neither the minimal nor the maximal tensor product.

My question: I wonder how much one can infer about the structure of the state space $\Omega_A \otimes \Omega_B$ from the structure of the local state spaces $\Omega_A$, $\Omega_B$ when considering different kinds of tensor products. More precisely, I wonder whether one can relate the statements that the local states form a polytope and that the composite states form a polytope (are there tensor products such that one statement implies the other?). Are there tensor products such that the composite states form a polytope while the local states do not? Are there tensor products such that the composite states always form a polytope? I am interested in all kinds of arguments that make statements about sets of states being (non-)polytopic when arising from certain kinds of tensor products.

I appreciate any kind of argument or comment, etc.

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Must Ω be a closed set? – Incnis Mrsi Aug 22 '14 at 16:44
Yes, usually, it is assumed to be a closed set. – Tom Jonathan Aug 22 '14 at 22:23

“Always a polytope” – definitely not. Moreover, in certain situation $Ω$, if a closed set, may not change at all; I mean product with the 0-dimensional set of states $Ω_{\rm id} = \{1\}$ (one point), considered as a subset of 1-dimensional vector space $V_{\rm id} = {\mathbb R}$. It has the only effect, the unit effect, and corresponds to 1-state quantum system.
Take the second system with $Ω_B\subset V_B$, and we see that $V_{\rm id}\otimes V_B = V_B$. Then, it is obvious to see that positivity defines a closed cone made of $Ω_B$, whereas normalization gives affine hyperplane $u^B = 1$ that contained all of $Ω_B$, so we have $Ω_B$ again. On the other hand, minimal tensor product is also $Ω_B$ by construction. No state or effect can be neither added nor removed.
Thus, the point set $Ω_{\rm id}$ is an identity object of any such tensor product definition, either maximal, or minimal, or whichever in the gap. Product of $Ω_{\rm id}$ and any second system with closed $Ω_2$ must give exactly this system: the same $Ω_2$ and, obviously, the same $E(Ω_2)$. But if to restrict OP’s question to products of more-than-0-dimensional $Ω$, then I do not know the answer.